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The self-inductance of a coaxial cable with inner-cylinder radius $a$ and outer-cylinder radius $b$, length $\ell$, carrying current $I$, is:
A$\dfrac{2\pi\mu_0 \ell}{\ln(b/a)}$
B$\dfrac{\mu_0 \ell}{2\pi}\,\ln(b/a)$
C$\dfrac{\mu_0 \ell}{2\pi}(b - a)$
D$\dfrac{\mu_0 \ell}{4\pi}(b^2 - a^2)$
Answer & Solution
Correct answer: B. $\dfrac{\mu_0 \ell}{2\pi}\,\ln(b/a)$
Between the cylinders $B = \mu_0 I/(2\pi r)$. Integrating energy density over the cylindrical shell volume yields $W = (\mu_0 I^2 \ell)/(4\pi) \cdot \ln(b/a)$. Equating to $\tfrac{1}{2}LI^2$ gives $L = \mu_0 \ell \ln(b/a)/(2\pi)$.
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